# Minimal surface in a ball

Assume a minimal surface $\Sigma$ has boundary on the unit sphere in the Euclidean space and $r$ is the distance from $\Sigma$ to the center of the ball. Is it true that $$\mathop{\rm area} \Sigma\ge \pi\cdot(1-r^2).$$

• If $r=0$, the statement follows directly from the monotonicity formula.
• If $\Sigma$ is topological disc the answer is YES, see answer of Oleg Eroshkin below.
• The general question is formulated as a conjecture in 1975 --- see comment of Ian Agol.
• There is an analog in all dimension and codimension for area minimizing surfaces, see Alexander, H.; Hoffman, D.; Osserman, R. Area estimates for submanifolds of Euclidean space. 1974.
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A little more background would be helpful. Why do you want to know this? –  Scott Morrison Nov 3 '09 at 18:02
This question is Conjecture 2 of: books.google.com/… –  Ian Agol Nov 22 '09 at 16:26

This result (and several similar) proved in a nice paper Alexander, H.; Osserman, R. "Area bounds for various classes of surfaces." Amer. J. Math. 97 (1975), no. 3, 753--769.

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Only for disc... but that is already a lot. !!!THANK YOU VERY MUCH!!! –  Anton Petrunin Nov 15 '09 at 4:56
One obvious observation (of which you are probably already aware) is that if the boundary of the surface is connected, it must have length at least $2\pi\sqrt{1-r^2}$, or else it is contained in a lune whose convex hull does not contain a point at distance $r$ from the center. In the very special case that your surface is a topological disk transverse to a foliation of the ball by concentric spheres, the coarea formula (obtained by integrating the lengths of the intersection of your surface with concentric spheres, and using this observation) gives an estimate for the area, but a quick calculation shows that it is not good enough to prove what you want.
It seems you only use that $\Sigma$ is saddle. This way you will only get that length $\ge 4\cdot\arccos r$ and you can not make it better. –  Anton Petrunin Nov 13 '09 at 13:30
(1) Did you meant that you have example for all small $a$? - (2) guess you mean rotation of $y=a\cosh ax$ around $x$-axis? Then $r=a$, but not $2a$(?) (3) In the very end, $2(\ln a)^2$ is huge number while $\pi(1-4a^2)$ is small --- not other way arround (?) (4) I see that as $a\to0$ the we have that $area-\pi\cdot(1-a^2)$ is very small, but it seems that it is still positive... –  Anton Petrunin Nov 12 '09 at 4:27